The union-closed sets conjecture almost holds for almost all random bipartite graphs
aa r X i v : . [ m a t h . C O ] F e b The union-closed sets conjecture almost holds foralmost all random bipartite graphs
Henning Bruhn and Oliver Schaudt
Abstract
Frankl’s union-closed sets conjecture states that in every finite union-closed set of sets, there is an element that is contained in at least half of themember-sets (provided there are at least two members). The conjecturehas an equivalent formulation in terms of graphs: In every bipartite graphwith least one edge, both colour classes contain a vertex belonging to atmost half of the maximal stable sets.We prove that, for every fixed edge-probability, almost every randombipartite graph almost satisfies Frankl’s conjecture.
One of the most basic conjectures in extremal set theory is Frankl’s conjectureon union-closed set systems. A set X of sets is union-closed if X ∪ Y ∈ X forall X, Y ∈ X . Union-closed sets conjecture.
Let
X 6 = {∅} be a finite union-closed set ofsets. Then there is a x ∈ S X ∈X X that lies in at least half of the members of X . While Frankl [9] dates the conjecture to 1979, it apparently did not appearin print before 1985, when it was mentioned as an open problem in Rival [17].Despite being widely known, there is only little substantial progress on theconjecture.The conjecture has two equivalent formulations, one in terms of lattices andone in terms of graphs. For the latter, let us say that a vertex set S in a graphis stable if no two of its vertices are adjacent, and that it is maximally stable if,in addition, every vertex outside S has a neighbour in S . Conjecture 1 (Bruhn, Charbit and Telle [4]) . Let G be a bipartite graph withat least one edge. Then each of the two bipartition classes contains a vertexbelonging to at most half of the maximal stable sets. We prove a slight weakening of Conjecture 1 for random bipartite graphs.For δ >
0, we say that a bipartite graph satisfies the union-closed sets conjectureup to δ if each of its two bipartition classes has a vertex for which the numberof maximal stable sets containing it is at most + δ times the total number1f maximal stable sets. A random bipartite graph is a graph on bipartitionclasses of cardinalities m and n , where any two vertices from different classesare independently joined by an edge with probability p . We say that almostevery random bipartite graph has property P if for every ε > N such that, whenever m + n ≥ N , the probability that a random bipartite graphon m + n vertices has P is at least 1 − ε .We prove: Theorem 2.
Let p ∈ (0 , be a fixed edge-probability. For every δ > , almostevery random bipartite graph satisfies the union-closed sets conjecture up to δ . While Frankl’s conjecture has attracted quite a lot of interest, a proof seemsstill out of reach. For a fairly complete bibliography on the conjecture, seeCz´edli, Mar´oti and Schmidt [6].Some of the earliest results verified the conjecture for few sets or few ele-ments in the ground set, that is, when n = |X | or m = | S X ∈X X | are small.The current best results show that the conjecture holds for m ≤
11, which is dueto Boˇsnjak and Markovi´c [3], and for n ≤
46, proved by Nishimura and Taka-hashi [13] and independently Roberts and Simpson [18]. The conjecture is alsoknown to be true when n is large compared to n , that is n ≥ m − √ m (Nishimura and Takahashi [13]). The latter result was improved upon byCz´edli [5], who shows that n ≥ m − √ m is enough. Recently, Balla, Bol-lobas and Eccles [1] pushed this to n ≥ ⌈ m +1 ⌉ .The lattice formulation of the conjecture was apparently known from veryearly on, as it is already mentioned in Rival [17]. Poonen [14] investigatedseveral variants and gave proofs for geometric as well as distributive lattices.Reinhold [16] extended this, with a very concise argument, to lower semimodularlattices. Finally, the conjecture holds as well for large semimodular lattices andfor planar semimodular lattices (Cz´edli and Schmidt [7]).The third view, in terms of graphs, on the union-closed sets conjectureis more recent. So far, the graph formulation is only verified for chordal-bipartite graphs and for bipartitioned circular interval graphs (Bruhn, Charbitand Telle [4]).One of the main techniques that is used for the set formulation of Frankl’sconjecture as well as for the lattice formulation, is averaging: The average fre-quency of an element is computed, and if that average is at least half of the sizeof the set system, it is concluded that the conjecture holds for the set system.Averaging is also our main tool. We discuss averaging and its limits in Section 3. In our graph-theoretic notation we usually follow Diestel [8], while we refer toBollob´as [2] for more details on random graphs.All our graphs are finite and simple. We always consider a bipartite graph G to have a fixed bipartition, which we denote by ( L ( G ) , R ( G )). When discussing2he bipartition classes, we will often refer to L ( G ) as the left side and to R ( G )as the right side of the graph.Throughout the paper we consider a fixed edge probability p with 0 < p < q = 1 − p . A random bipartite graph G is a bipartite graphwhere every pair u ∈ L ( G ) and v ∈ R ( G ) is joined by an edge independently withprobability p . We denote by B ( m, n ; p ) the probability space whose elementsare the random bipartite graphs G with | L ( G ) | = m and | R ( G ) | = n . We willalways tacitly assume that m ≥ n ≥
1. Indeed, if one of the sides of therandom bipartite graph is empty, then the graph has no edge and is thereforetrivial with respect to Conjecture 1.Markov’s inequality states that for a non-negative random variable X andany α >
0, Pr[ X ≥ α ] ≤ E[ X ] α . (1)Chebyshev’s inequality is as follows. Let X be a random variable with finitevariance σ = E[ X ] − E[ X ] . Then, for every real λ > | X − E[ X ] | ≥ λ ] ≤ σ λ . (2) Most of the partial results on Frankl’s conjecture are based on one of two tech-niques:
Local configurations and averaging . By a local configuration we mean asubsystem of the union-closed set system X , that guarantees that one elementof the ground set lies in at least half of the members of X . For instance, oneof the earliest results is the observation of Sarvate and Renaud [19] that theelement of a singleton will always belong to at least half of the sets. More localconfigurations have later been found by Poonen [14], Vaughan [21], Morris [12]and others.The second technique consists in taking the average of the number of membersets containing a given element, where the average ranges over the set U = S X ∈X X of all elements. If that average is at least | U | then clearly X willsatisfy the conjecture. Averaging was used successfully by Cz´edli [5] to showthat the conjecture holds when there are vastly more sets than elements, thatis, when |X | ≥ | U | − √ | U | . Reimer [15] showed that the average is always atleast log ( | U | ).Averaging will not always work. It is easy to construct union-closed setsystems in which the average is too low. Cz´edli, Mar´oti and Schmidt [6] evenfound such set systems of size |X | = ⌊ | U | +1 / ⌋ . Nevertheless, we will see that,in the graph formulation, averaging will almost always allow us to conclude thatthe union-closed sets conjecture is satisfied (up to any δ > A ( G )for the set of maximal stable sets of a bipartite graph G . The graph formulationof the union-closed sets conjecture, Conjecture 1, is satisfied if G contains an3 nstable vertex in both bipartition classes, that is, a vertex that lies in at mosthalf of the maximal stable sets. We note first that exchanging the sides turnsa random bipartite graph G ∈ B ( m, n ; p ) into a member of B ( n, m ; p ), whichmeans that it will suffice to show the existence of an unstable vertex in L ( G ).All the discussion and proofs that follows will focus on the left side L ( G ).That a vertex v is unstable means that |A v ( G ) | , the number of maximalstable sets containing v , is at most |A ( G ) | . Thus, if for the average X v ∈ L ( G ) |A v ( G ) ||A ( G ) | ≤ | L ( G ) | then L ( G ) will contain an unstable vertex. Double-counting shows that theabove average is equal toleft-avg( G ) := X A ∈A ( G ) | A ∩ L ( G ) ||A ( G ) | , and thus our aim is to show that when m + n is very large, it follows with highprobability that left-avg( G ) ≤ m for any G ∈ B ( m, n ; p ).Unfortunately, we will not reach this aim. While we will show for large partsof the parameter space ( m, n ) that the average is, with high probability, smallenough, we will also see that when n is roughly q − m the average becomes veryclose to m , so close that our tools are not sharp enough to separate the averagefrom slightly above m . Therefore, we provide for a bit more space by settlingon bounding the average away from ( + δ ) m for any positive δ , which then onlyallows us to deduce the existence of a vertex v ∈ L ( G ) that is almost unstable ,in the sense that v lies in at most ( + δ ) |A ( G ) | maximal stable sets.Much of the previous discussion is subsumed in the following lemma. Lemma 3.
Let G be a bipartite graph, and let δ ≥ . If left-avg( G ) ≤ (cid:0) + δ (cid:1) | L ( G ) | then there exists a vertex in L ( G ) that lies in at most (cid:0) + δ (cid:1) |A ( G ) | maximalstable sets.Proof. Double counting yields P y ∈ L ( G ) |A y ( G ) | = P A ∈A ( G ) | A ∩ L ( G ) | , fromwhich we deduce that P y ∈ L ( G ) |A y ( G ) | ≤ | L ( G ) | · (cid:0) + δ (cid:1) |A ( G ) | . Thus there isa y ∈ L ( G ) with |A y ( G ) | ≤ (cid:0) + δ (cid:1) |A ( G ) | .Most of the effort in this article will be spent on proving the following result,which is the heart of our main result, Theorem 2: Theorem 4.
For all δ > and all ε > there is an integer N so that for G ∈ B ( m, n ; p ) Pr (cid:2) left-avg( G ) ≤ (cid:0) + δ (cid:1) m (cid:3) ≥ − ε for all m, n with m + n ≥ N and n ≥ max { , ( ⌈ /q (2) ⌉ + 2) } + 1 .
4n order to show how Theorem 2 follows from Theorem 4, we need to dealwith the special case when one side is of constant size while the other becomesever larger. Indeed, in this case averaging might fail—for a trivial reason. If wefix a constant right side R ( G ), while L ( G ) becomes ever larger, then L ( G ) willcontain many isolated vertices. Since the isolated vertices lie in every maximalstable set they may push up left-avg( G ) to above m .However, isolated vertices are never a threat to Frankl’s conjecture: A bi-partite graph satisfies the union-closed sets conjecture if and only if it satisfiesthe conjecture with all isolated vertices deleted. More generally, it turns outthat the special case of a constant right side is easily taken care of: Lemma 5.
Let c be a positive integer, and let ε > . Then there is an N sothat for G ∈ B ( m, n ; p )Pr [ L ( G ) contains an unstable vertex] ≥ − ε, for all m, n with m ≥ N and n ≤ c .Proof. Let G be any bipartite graph, and suppose there is a vertex v ∈ L ( G )that is adjacent with every vertex in R ( G ). Then, the only maximal stable setthat contains v is L ( G ). Since the fact that v is incident with an edge impliesthat G has at least two maximal stable sets, v is unstable.We now calculate the probability that there is such a vertex. The probabilitythat R ( G ) = N ( v ) for a fixed vertex v ∈ L ( G ) is p n ≥ p c if n ≤ c . Thus theprobability that no such vertex exists in L ( G ) is at most (1 − p c ) m , which tendsto 0 as m → ∞ . Proof of Theorem 2.
Let δ > L ( G ) of almost every random bipartite graph G in B ( m, n ; p )contains a vertex that lies in at most ( + δ ) |A| maximal stable sets. For this,consider a ε >
0, and let N be the maximum of the N given by Theorem 4 andLemma 5 with c = max { , ( ⌈ /q (2) ⌉ +2) } . Consider a pair m, n of positiveintegers with m + n ≥ N . If n ≤ max { , ( ⌈ /q (2) ⌉ + 2) } then Lemma 5yields an unstable vertex in L ( G ) with probability at least 1 − ε . If, on the otherhand, n ≥ max { , ( ⌈ /q (2) ⌉ + 2) } + 1, Theorem 4 becomes applicable,which is to say that with probability at least 1 − ε we have left-avg( G ) ≤ (cid:0) + δ (cid:1) m . Now, Lemma 3 yields the desired vertex in L ( G ).We close this section with the obvious but useful observation that if thereare many more maximal stable sets with small left side than with large leftside, then the average over the left sides is small, too. We will use this lemmarepeatedly. Lemma 6.
Let ν > and δ ≥ , and let G be a bipartite graph with | L ( G ) | = m .Let L be the maximal stable sets A of G with | A ∩ L ( G ) | ≥ ( + δ ) m , and let S be those maximal stable sets B with | B ∩ L ( G ) | ≤ (1 − ν ) m . If |S| ≥ ν |L| then left-avg( G ) ≤ (cid:0) + δ (cid:1) m. roof. Let M = A ( G ) \ ( L ∪ S ), that is, M is the set of those maximal stablesets A with (1 − ν ) m < | A ∩ L ( G ) | < ( + δ ) m . Then X A ∈A ( G ) | A ∩ L ( G ) | (cid:0) + δ (cid:1) m ≤ X A ∈L m (cid:0) + δ (cid:1) m + X B ∈M ( + δ ) m (cid:0) + δ (cid:1) m + X C ∈S (1 − ν ) m (cid:0) + δ (cid:1) m ≤ |L| + |M| + (1 − ν ) |S| = |A ( G ) | − ( ν |S| − |L| ) . Thus, from |S| ≥ ν |L| it follows that P A ∈A ( G ) | A ∩ L ( G ) | ( + δ ) m ≤ |A ( G ) | , which isequivalent to the inequality of the lemma. In order to prove Theorem 4, we distinguish several cases, depending on therelative sizes, m and n , of the two sides of the random bipartite graph G ∈B ( m, n ; p ). In each of the cases we need a different method.The general strategy follows Lemma 6: We bound the number of maximalstable sets with large left side, usually counted by a random variable L G , andat the same time we show that there are many maximal stable sets with a smallleft side; those we count with S G .Up to n < q − m we are able to use the same bound for the number L G ofmaximal stable sets whose left sides are of size at least m : We prove that withhigh probability L G is bounded by a polynomial in n . For right sides that aremuch larger than the left side, i.e. m ≫ n , we even extend such a bound tomaximal stable sets with left side ≥ m .For the maximal stable sets with small left side, counted by S G , we need todistinguish several cases. When the left side of the graph is much larger thanthe right side, namely m ≥ q − √ n , we find with high probability a large inducedmatching in G . This in turn implies that the total number of maximal stablesets is high, and thus clearly also the number of those with small left side.When the sides of the graph do not differ too much in size, m ≤ q − √ n and n ≤ q − √ m , the variance of the number of maximal stable sets with small leftside is moderate enough to apply Chebychev’s inequality. Since the expectationof S G is high, we again can use Lemma 6 to deduce Theorem 4.However, when the left side of the graph becomes much larger than theright side, we cannot control the variance of S G anymore. Instead, for q − √ m ≤ n ≤ q − m , we cut the right side into many pieces each of large size and applyHoeffding’s inequality to each of the pieces together with the left side. Theinequality ensures that we find on at least one of the pieces a large number ofmaximal stable sets of small left side. Surpassing n ≥ q − m , we have to refineour estimations but we can still use this strategy up to slightly below n = q − m .In the interval q − m ≤ n ≤ q − m , we encounter a serious obstacle. There,we have to cope with an average that is very close to m . It is precisely forthis reason that, overall, we only prove that left-avg( G ) ≤ (cid:0) + δ (cid:1) m insteadof left-avg( G ) ≤ m . To keep below the slightly higher average, we only need6o bound the number of maximal stable sets with left side > ( + δ ) m . Thisnumber we will almost trivally bound by 2 λm , with some λ <
1. On the otherhand, we will see that the number S G of maximal stable sets of small left sideis 2 λ ′ m with a λ ′ as close to 1 as we want.In the remaining case, we are dealing with an enormous right side: n ≥ q − m .Then, it is easy to see that with high probability there is an induced matchingthat covers all of the left side, which implies that every subset of L ( G ) is theleft side of a maximal stable set. This immediately gives us left-avg( G ) = m . m ≥ q − √ n In this section we treat the graphs whose left side is much larger than the rightside. From an easy argument it follows that, with high probability, any largeenough random graph G ∈ B ( m, n ; p ) contains an induced matching of size s log ( n ), for any constant s . This directly implies that the total number ofmaximal stable sets is large. At the same time, we shall bound the number ofmaximal stable sets of large left side, which then shows that there are many ofsmall left side.However, if we take large left side to mean at least m then it might be thatmost of those of small left side have a left side whose size is very close to m .Such a left side does not help much to drop the average. Therefore, we willconsider a more generous notion of a large left side and bound the number ofmaximal stable sets that have a left side of ≥ m ; then small will mean < m .For a random graph G ∈ B ( m, n ; p ), let stab( ≥ ℓ ; ≥ r ) denote the numberof stable sets of that have at least ℓ vertices in L ( G ) and at least r verticesin R ( G ). Lemma 7.
Let ℓ ∗ ≤ m and r ∗ ≤ n so that nq ℓ ∗ ≤ . Then for G ∈ B ( m, n ; p )E[ stab ( ≥ ℓ ∗ ; ≥ r ∗ )] ≤ m +1 (cid:16) nq ℓ ∗ (cid:17) r ∗ . Proof.
The expectation is given byE[stab( ≥ ℓ ∗ ; ≥ r ∗ )] = m X ℓ = ℓ ∗ n X r = r ∗ (cid:18) mℓ (cid:19)(cid:18) nr (cid:19) q ℓr ≤ m X ℓ = ℓ ∗ (cid:18) mℓ (cid:19)! n X r = r ∗ (cid:18) nr (cid:19) q ℓ ∗ r ! ≤ m n X r = r ∗ (cid:16) nq ℓ ∗ (cid:17) r ! ≤ m ∞ X r = r ∗ (cid:16) nq ℓ ∗ (cid:17) r ! . The error estimation for the geometric series yields P ∞ i = k z i ≤ | z | k for any | z | ≤ . Applying this for z = nq ℓ ∗ , we obtain the claimed bound of thelemma. 7n the following, we denote by L ′ G the number of maximal stable sets S of abipartite graph G with | S ∩ L ( G ) | ≥ m . Lemma 8.
Let r ∗ = ⌈ /q (2) ⌉ + 1 . Then for any ε > there is an N so thatfor G ∈ B ( m, n ; p ) Pr[ L ′ G ≤ n r ∗ ] ≥ − ε for all m, n with m ≥ q − √ n and m + n ≥ N .Proof. Throughout the proof we assume that m ≥ q − √ n .Setting ℓ ∗ = m , we get from Lemma 7 thatE[stab( ≥ m ; ≥ r ∗ )] ≤ m +1 (cid:0) nq m (cid:1) r ∗ = 2 n r ∗ · q m ( r ∗ − log /q (2)) ≤ n r ∗ · q m , by choice of r ∗ .Choose N so that 2 n r ∗ · q m ≤ ε for all m, n with m + n ≥ N . Then, fromMarkov’s inequality it follows thatPr[stab( ≥ m ; ≥ r ∗ ) > ≤ ε. (3)The set of maximal stable sets S of G whose left side S ∩ L ( G ) has size atleast m is divided into those S with | S ∩ R ( G ) | ≥ r ∗ and those whose right sideshave < r ∗ vertices; let the number of the latter ones be t . Since there are atmost n r ∗ subsets of R ( G ) with at most r ∗ vertices, t ≤ n r ∗ . Hence, L ′ G ≤ stab( ≥ m ; ≥ r ∗ ) + t ≤ stab( ≥ m ; ≥ r ∗ ) + n r ∗ . From (3), we deduce Pr[ L ′ G > n r ∗ ] ≤ ε . Lemma 9.
Let s be a positive integer, and let ε > . Then there is an N sothat for G ∈ B ( m, n ; p )Pr [ G has an induced matching of size ≥ s log ( n )] ≥ − ε for all m, n with m + n ≥ N , n ≥ s log ( n ) and m ≥ q − √ n .Proof. Let us assume throughout the proof that n ≥ s log ( n ) and m ≥ q − √ n .Put k := ⌈ s log ( n ) ⌉ , and choose ⌊ m/k ⌋ pairwise disjoint subsets L , . . . , L ⌊ m/k ⌋ of size k of L ( G ). Since n ≥ s log ( n ), n ≥ k and so we may choose a set R ′ ⊆ R ( G ) with | R ′ | = k . For i = 1 , . . . , ⌊ m/k ⌋ let M i be the random indicatorvariable for an induced matching of size k on L i ∪ R ′ . It is straightforward thatPr [ M i = 1] = k ! p k q k − k ≥ p k q k . Since the M i are independent,Pr ⌊ m/k ⌋ X i =1 M i = 0 ≤ (cid:16) − p k q k (cid:17) ⌊ m/k ⌋ ≤ e − p k q k ⌊ m/k ⌋ , − x ≤ e x for all x <
1. Now for large m + n the dominating term in p k q k ⌊ m/k ⌋ is q k m , since m ≥ q − √ n , which becomesarbitrarily large for large m + n as k = ⌈ s log ( n ) ⌉ and n ≤ (cid:16) log /q ( m ) (cid:17) . Thus,there is an N so that Pr hP ⌊ m/k ⌋ i =1 M i = 0 i ≤ ε for all m, n with m + n ≥ N .We have now bounded the number L ′ G of maximal stable sets of large leftside, while the previous lemma will let us to conclude that the number of thosewith small left side is large. Together this allows us prove the first case ofTheorem 4: Lemma 10.
For every ε > there exists an N so that for G ∈ B ( m, n ; p )Pr (cid:2) left-avg( G ) ≤ m (cid:3) ≥ − ε, for all m, n with m + n ≥ N , n ≥ max { , ( ⌈ /q (2) ⌉ + 2) } and m ≥ q − √ n .Proof. Set r ∗ = ⌈ /q (2) ⌉ + 1. Choose N to be the maximum of the N obtained from Lemma 8 for ε and the one from Lemma 9 applied to s = r ∗ +1 = ⌈ /q (2) ⌉ + 2 and ε .Now, consider m, n with m + n ≥ N , n ≥ max { , ( ⌈ /q (2) ⌉ + 2) } and m ≥ q − √ n . We note that n ≥ max { , ( ⌈ /q (2) ⌉ + 2) } implies that n ≥ s log ( n ). By choice of s and N , we obtain from Lemma 9 that the probabilitythat G ∈ B ( m, n ; p ) does not contain an induced matching of size at least s log ( n ) is at most ε . On the other hand, the probability that the number L ′ G of maximal stable sets A with | A ∩ L ( G ) | ≥ m surpasses n r ∗ is as well ≤ ε .Thus, the probability that none of these two events occur is at least 1 − ε . Weclaim that in this case left-avg( G ) ≤ m .So, assume that G contains an induced matching of cardinality ≥ s log ( n )and that L ′ G ≤ n r ∗ . There are at least 2 s log ( n ) maximal stable sets on thesubgraph restricted to the matching edges. Since each extends to a distinctmaximal stable set of G , the number of maximal stable sets of G is at least2 s log ( n ) = n s = n r ∗ +1 . On the other hand, from L ′ G ≤ n r ∗ it follows thatat least n r ∗ ( n −
1) of the maximal stable sets have a left side of size at most m . As n − ≥
3, we may apply Lemma 6 with δ = 0 in order to see thatleft-avg( G ) ≤ m .The key observation in the argument above is that the number of maximalstable sets with a large left side is bounded by a polynomial in n , the size ofthe right-hand side. We will continue to exploit this, in a slightly strengthenedversion, below. The second part of the argument here is to note that thereis always a relatively large induced matching, from which we deduce that thetotal number of maximal stable sets is not too small. Then we also have a largenumber of maximal stable sets with small left side, so that we are guaranteed asmall average.This strategy fails once m becomes smaller than n . Assume m < n and, forsimplicity, p = q = . Below we will in that case bound the number of maximal9table sets with large left side by about 2 n . Thus, for our strategy to work,we should better find an induced matching of size at least 2 log ( n ). An easycalculation, however, shows that the expected number of induced matchings ofsize 2 log ( n ) is below one. n ≤ q − √ m and m ≤ q − √ n From now on we will denote by L G the number of maximal stable sets S with | S ∩ L ( G ) | ≥ m of a random bipartite graph G ∈ B ( m, n ; p ). We first bound L G by a polynomial in n , a bound that will be useful up to slightly below n = 2 m . Lemma 11.
For every α < and every ε > there exists an N so that for G ∈ B ( m, n ; p ) Pr[ L G ≤ n log q (1 / ] ≥ − ε when m + n ≥ N and n ≤ q − αm .Proof. Let α < be given and assume n ≤ q − αm .We determine first the probability that a random bipartite graph containsmany stable sets (not necessarily maximal) with left side ≥ m and right side ≥ ⌊ log q (1 / ⌋ + 1.For this, note that α < implies nq m ≤ q m ( − α ) ≤ for large m . Moreover,it follows that ν := (1 / − α )( ⌊ log q (1 / ⌋ + 1 − log q (1 / > . Thus, applying Lemma 7 yieldsE[stab( ≥ m ; ≥ ⌊ log q (1 / ⌋ + 1)] ≤ m +1 (cid:0) nq m (cid:1) ⌊ log q (1 / ⌋ +1 = 2 n log q (1 / m n ⌊ log q (1 / ⌋ +1 − log q (1 / q ⌊ log q (1 / ⌋ +12 m ≤ n log q (1 / q − log q (1 / m − α ( ⌊ log q (1 / ⌋ +1 − log q (1 / m + ⌊ log q (1 / ⌋ +12 m = 2 n log q (1 / q m (1 / − α )( ⌊ log q (1 / ⌋ +1 − log q (1 / ≤ n log q (1 / q νm for sufficiently large m . With Markov’s inequality we deducePr[stab( ≥ m ; ≥ ⌊ log q (1 / ⌋ + 1) > n log q (1 / ] ≤ n log q (1 / q νm n log q (1 / = 2 q νm , which tends to 0 as m → ∞ . Since n ≤ q − αm implies that also m must be largefor large m + n , we may find an N so that Pr[stab( ≥ m ; ≥ ⌊ log q (1 / ⌋ + 1) >n log q (1 / ] ≤ ε, for all for all integers m, n with m + n ≥ N .Considering such m and n , we turn now to the number of maximal stablesets L G with left side ≥ m . As in Lemma 8, we argue that the maximal stable10ets counted by L G split into those whose right side have at least ⌊ log q (1 / ⌋ + 1vertices and those with at most ⌊ log q (1 / ⌋ vertices in R ( G ). Of the latter ones,there are at most n log q (1 / many sets. By choice of N , the probability that wehave more than n log q (1 / of the former is bounded by ε .Let us quickly calculate the probability that a given set of vertices is amaximal stable set. Lemma 12.
For a random bipartite graph G ∈ B ( m, n ; p ) , let S be a subset of V ( G ) . If | S ∩ L | = ℓ and | S ∩ R | = r then Pr[ S ∈ A ] = q ℓr (1 − q r ) m − ℓ (cid:0) − q ℓ (cid:1) n − r . Proof.
The factor q ℓr is the probability that there is no edge from S ∩ L ( G ) to S ∩ R ( G ), that is, S is a stable set. The factor (1 − q r ) m − ℓ is the probabilitythat every of the m − ℓ many vertices in L ( G ) \ S has a neighbour in S ∩ R ( G ),and (cid:0) − q ℓ (cid:1) n − r is the probability that every of the n − r many vertices in R ( G ) \ S has a neighbour in S ∩ L ( G ). The latter two conditions ensure that S is a maximal stable set.Next, we calculate the expectation and the variance of the number of maxi-mal stable sets of small left side. Since they outnumber the other maximal stablesets by far, we concentrate on those maximal stable sets with a left side equalto ≈ log /q ( n ) and a right side equal to ≈ log /q ( m ). This choice is somewhatforced by the maximality requirement for maximal stable sets: For logarithmicsized left sides the maximality requirement eliminates only a constant propor-tion of the possible maximal stable sets. With smaller sides, on the other hand,we lose more sets, that is, the expectation becomes much smaller.For G ∈ B ( m, n ; p ) we denote by S G the number of maximal stable sets S of G with | S ∩ L ( G ) | = ⌊ log /q ( n ) ⌋ and | S ∩ R ( G ) | = ⌊ log /q ( m ) ⌋ . Lemma 13.
Let c = e − (2 /q +1) . There are m , n ∈ N such that for G ∈B ( m, n ; p ) E[ S G ] ≥ c (cid:18) m ⌊ log /q ( n ) ⌋ (cid:19) ⌊ log /q ( m ) ⌋ −⌊ log /q ( m ) ⌋ for all m ≥ m , n ≥ n with m ≥ log /q ( n ) and n ≥ log /q ( m ) .Proof. Assume that m, n are integers with m ≥ log /q ( n ) and n ≥ log /q ( m ).We first note that (cid:16) − q ⌊ log /q ( m ) ⌋ (cid:17) m −⌊ log /q ( n ) ⌋ ≥ (cid:16) − q ⌊ log /q ( m ) ⌋ (cid:17) m ≥ (cid:16) − q log /q ( m ) − (cid:17) m = (cid:18) − qm (cid:19) m . m →∞ (cid:16) − qm (cid:17) m = e − /q , there is m ∈ N such that (cid:18) − qm (cid:19) m ≥ e − (cid:16) q + 12 (cid:17) , (4)for all m ≥ m . With the same arguments, we see that there is an n ∈ N , sothat (cid:16) − q ⌊ log /q ( n ) ⌋ (cid:17) n −⌊ log /q ( m ) ⌋ ≥ e − (cid:16) q + 12 (cid:17) . for all n ≥ n .Now consider a random bipartite graph G ∈ B ( m, n ; p ) with m ≥ m and n ≥ n , and let S be any vertex subset with | S ∩ L ( G ) | = ⌊ log /q ( n ) ⌋ =: a and | S ∩ R ( G ) | = ⌊ log /q ( m ) ⌋ =: b . By Lemma 12, the probability that S is amaximal stable set of G amounts toPr[ S is maximally stable] = q ab (cid:0) − q b (cid:1) m − a (1 − q a ) n − b . The first term is at least equal to n −⌊ log /q ( m ) ⌋ , while, by (4), the remainingterms together are at least equal to c = e − (2 /q +1) . This yieldsPr[ S is maximally stable] ≥ cn −⌊ log /q ( m ) ⌋ . Thus, E[ S G ] ≥ (cid:18) m ⌊ log /q ( n ) ⌋ (cid:19)(cid:18) n ⌊ log /q ( m ) ⌋ (cid:19) cn −⌊ log /q ( m ) ⌋ ≥ (cid:18) m ⌊ log /q ( n ) ⌋ (cid:19) n ⌊ log /q ( m ) ⌋ ! ⌊ log /q ( m ) ⌋ cn −⌊ log /q ( m ) ⌋ = c (cid:18) m ⌊ log /q ( n ) ⌋ (cid:19) ⌊ log /q ( m ) ⌋ −⌊ log /q ( m ) ⌋ . We use Chebyshev’s inequality to show that, with high probability, S G doesnot differ much from the expected value. Lemma 14.
For every ε > there is an N so that for G ∈ B ( m, n ; p )Pr (cid:2) S G > E[ S G ] (cid:3) ≥ − ε, for all m, n with m + n ≥ N , m ≤ q − √ n and n ≤ q − √ m .Proof. Since the statement of Lemma 14 is symmetric in m and n , we mayassume throughout the proof that m ≤ n . Moreover we assume that m ≤ q − √ n and n ≤ q − √ m . 12et a := ⌊ log /q ( n ) ⌋ and b := ⌊ log /q ( m ) ⌋ . Chebyshev’s inequality (2) givesus Pr (cid:2) S G ≤ E[ S G ] (cid:3) ≤ σ E[ S G ] , where σ = E[ S G ] − E[ S G ] is the variance of the random variable S G . We haveE[ S G ] = a X i =0 b X j =0 A i,j , where A i,j denotes the expected number of pairs ( S, T ) of maximal stable sets of G with | S ∩ L ( G ) | = a = | T ∩ L ( G ) | , | S ∩ R ( G ) | = b = | T ∩ R ( G ) | , | S ∩ T ∩ L ( G ) | = i , and | S ∩ T ∩ R ( G ) | = j . By Lemma 12,E[ S G ] = (cid:18) ma (cid:19) (cid:18) nb (cid:19) q ab (1 − q a ) n − b ) (1 − q b ) n − a ) . (5)We will first show that there is an N so that A , − E[ S G ] E[ S G ] ≤ ε , (6)for all m, n with m + n ≥ N .To prove this, observe that A , ≤ (cid:18) ma (cid:19)(cid:18) m − aa (cid:19)(cid:18) nb (cid:19)(cid:18) n − bb (cid:19) q ab (1 − q a ) n − b ) (1 − q b ) m − a ) . Indeed, while the binomial coefficients count the number of possibilities tochoose the disjoint sets S and T , the factor q ab is the probability that S and T are stable sets. Furthermore, the probability that every vertex in R ( G ) \ ( S ∪ T )has a neighbour in S and a neighbour in T is equal to (1 − q a ) n − b ) ; the factor(1 − q b ) m − a ) expresses the analogous probability for L ( G ).The above estimation for A , together with (5) yields A , / E[ S G ] ≤ (1 − q a ) − b (1 − q b ) − a , and consequently A , − E[ S G ] E[ S G ] ≤ (1 − q a ) − b (1 − q b ) − a − . Next, note that if n is large enough so that nq ≤ then(1 − q a ) b ≥ (1 − q log /q ( n ) − ) /q ( m ) ≥ (cid:18) − nq (cid:19) √ n ≥ (cid:16) e − nq (cid:17) √ n → e = 1 as n → ∞ , where we have used that 1 − x ≥ e − x for all 0 ≤ x ≤ /
2. The analogousestimation holds for (1 − q b ) − a . Thus, if m and n are large enough, then(1 − q a ) − b (1 − q b ) − a − ≤ ε . m ≤ q − √ n and n ≤ q − √ m that both of m and n have tobe large if m + n is large, we may therefore choose N so that (6) holds.We will now investigate A i,j / E[ S G ] when i + j ≥
1. For this, let B i,j = (cid:18) mi (cid:19)(cid:18) m − ia − i (cid:19)(cid:18) m − aa − i (cid:19)(cid:18) nj (cid:19)(cid:18) n − jb − j (cid:19)(cid:18) n − bb − j (cid:19) q ab − ij . (7)Note that B i,j equals the expected number of pairs ( S, T ) of stable sets (notnecessarily maximal) with | S ∩ L ( G ) | = a = | T ∩ L ( G ) | , | S ∩ R ( G ) | = b = | T ∩ R ( G ) | , | S ∩ T ∩ L ( G ) | = i , and | S ∩ T ∩ R ( G ) | = j . Hence, A i,j ≤ B i,j forall 0 ≤ i ≤ a and 0 ≤ j ≤ b .For r, s ∈ N with r ≥ s , let ( r ) s denote the s -th falling factorial of r ,i.e., ( r ) s = r ( r − · · · ( r − s + 1). For the binomial coefficients appearing in B i,j / E[ S G ] that involve m , we deduce (cid:0) mi (cid:1)(cid:0) m − ia − i (cid:1)(cid:0) m − aa − i (cid:1)(cid:0) ma (cid:1) = ( m − a ) a − i ( a ) i ( m ) a i ! ≤ ( a ) i ( m ) i i ! ≤ ( a ) i m i ≤ a i m i , for all integers i with 0 ≤ i ≤ a . With the analogous estimation for the binomialcoefficients involving n , we obtain (cid:0) mi (cid:1)(cid:0) m − ia − i (cid:1)(cid:0) m − aa − i (cid:1)(cid:0) nj (cid:1)(cid:0) n − jb − j (cid:1)(cid:0) n − bb − j (cid:1)(cid:0) ma (cid:1) (cid:0) nb (cid:1) ≤ a i b j m i n j , (8)for all integers i, j with 0 ≤ i ≤ a and 0 ≤ j ≤ b .An easy calculation (very similar to (4)) shows that there is a constant c such that c ≥ − q a ) − n − b ) (1 − q b ) − m − a ) (9)for all m .Recalling the explicit expression (5) for E[ S G ] , and then applying first (8)and then (9) we deduce( a + 1)( b + 1) B i,j E[ S G ] = ( a + 1)( b + 1) (cid:0) mi (cid:1)(cid:0) m − ia − i (cid:1)(cid:0) m − aa − i (cid:1)(cid:0) nj (cid:1)(cid:0) n − jb − j (cid:1)(cid:0) n − bb − j (cid:1) q ab − ij (cid:0) ma (cid:1) (cid:0) nb (cid:1) q ab (1 − q a ) n − b ) (1 − q b ) m − a )(8) ≤ ( a + 1)( b + 1) a i b j m i n j q ij (1 − q a ) n − b ) (1 − q b ) m − a ) ≤ a i +1 b j +1 m i n j q ij (1 − q a ) n − b ) (1 − q b ) m − a )(9) ≤ ca i +1 b j +1 m i n j q ij for all i, j with i + j ≥
1, and where we assume in the third step that m and n are large enough so that a = ⌊ log /q ( n ) ⌋ ≥ b = ⌊ log /q ( m ) ⌋ ≥
1. (Again,14his is possible since m ≤ q − √ n and n ≤ q − √ m implies that both of m and n have to be large if m + n is large.)In order to continue with the estimation we consider the term m i n j q ij . For0 ≤ i ≤ a and 0 ≤ j ≤ b , we see that m i q ij ≥ ( mq b ) i ≥ ( mq log /q ( m ) ) i =( mm ) i = 1 . In a similar way, we obtain n j q ij ≥
1. Now, if i ≥ j then m i n j q ij = m i ( n j q ij ) ≥ m i . If, on the other hand, i < j then n j m i q ij ≥ n j ≥ m j , since m ≤ n . Thus m i n j q ij ≥ m max( i,j ) , for all m, n with m ≤ n. (10)Using (10), we obtain( a + 1)( b + 1) B i,j E[ S G ] ≤ ca i +1 b j +1 m i n j q ij ≤ ca i +1 b j +1 m max( i,j ) ≤ c log /q ( n ) i +1 log /q ( m ) j +1 m max( i,j ) ≤ c ( √ m ) i +1 log /q ( m ) j +1 m max( i,j ) ≤ cm i + log /q ( m ) j +1 m max( i,j ) , since n ≤ q − √ m and m ≤ n . Now, since i + j ≥
1, the last term tends to 0for m → ∞ . Therefore, there is an N independent of i, j such that for all m + n ≥ N ( a + 1)( b + 1) B i,j E[ S G ] ≤ ε , whenever i + j ≥ N = max( N , N ) and m, n with m + n ≥ N , we get with (6)Pr[ S G ≤ E[ S G ]] ≤ σ E[ S G ] ≤ A , − E[ S G ] E[ S G ] + 4( a + 1)( b + 1) max { B i,j : 0 ≤ i ≤ a, ≤ j ≤ b, i + j ≥ } E[ S G ] ≤ ε ε ε. Let us quickly explain why the method of Lemma 14 ceases to work when n ≥ q −√ m . In the proof we aim for Pr[ S G < E[ S G ]] to tend to 0 with growing m + n . We achieve that by forcing the right hand side of (8) to vanish for large m + n , and all i, j with i + j ≥
1. In particular, when i = 1 and j = 0, weneed a m = ⌊ log /q ( n ) ⌋ /m to vanish with growing m , which in turn requiresthat n < q −√ m . 15o finish this case, we observe that, with high probability, Lemma 11 boundsthe number L G of maximal stable sets with large left side with a polynomialin n , while we will see below that, again with high probability, Lemmas 13and 14 translate into superpolynomially many maximal stable sets with smallleft side. Lemma 15.
For every ε > there is an N so that for every G ∈ B ( m, n ; p )Pr[left-avg( G ) ≤ m ] ≥ − ε for all m, n with m + n ≥ N , m ≤ q − √ n and n ≤ q − √ m .Proof. Choose N large enough so that it is at least as large as the N inLemma 13 with ε and as the N in Lemma 14, as well with ε , and so that ⌊ log /q ( n ) ⌋ ≥ log /q ( n ) for any m, n with m + n ≥ N and m ≤ q − √ n .In the remainder of the proof, we consider integers m, n with m + n ≥ N , m ≤ q − √ n and n ≤ q − √ m . By Lemmas 13 and 14, there is a constant c > m and n , so that the probability that S G ≤ c m ⌊ log /q ( n ) ⌋ ! ⌊ log /q ( n ) ⌋ ⌊ log /q ( m ) ⌋ −⌊ log /q ( m ) ⌋ ≤ c (cid:18) m ⌊ log /q ( n ) ⌋ (cid:19) ⌊ log /q ( m ) ⌋ −⌊ log /q ( m ) ⌋ , is at most ε .Now, using that log /q ( m ) ≤ √ n and log /q ( n ) ≤ √ m we obtain c m ⌊ log /q ( n ) ⌋ ! ⌊ log /q ( n ) ⌋ ⌊ log /q ( m ) ⌋ −⌊ log /q ( m ) ⌋ ≥ cm log /q ( n ) · log /q ( n ) − log /q ( n ) · log /q ( m ) − log /q ( m ) ≥ cn log /q ( m ) · n − log /q (log /q ( n )) · (cid:0) √ n (cid:1) − log /q ( m ) ≥ cn log /q ( m ) · n − log /q ( √ m ) · ( n ) − log /q ( m ) = cn log /q ( m ) . It follows that Pr h S G ≤ cn log /q ( m ) i ≤ ε . (11)On the other hand, we obtain from Lemma 11 that there is a N so thatPr h L G > n log q (1 / i ≤ ε , (12)whenever m + n ≥ N .Recall that S G is a lower bound on the number of maximal stable sets S with | S ∩ L ( G ) | = ⌊ log /q ( n ) ⌋ . Since m ≤ q − √ n and n ≤ q − √ m implies thatboth of m and n have to be large if m + n is large, we may choose N large16nough so that √ m ≤ m and cn log /q ( m ) ≥ n log q (1 / whenever m + n ≥ N .We claim that if S G ≥ cn log /q ( m ) and L G ≤ n log q (1 / then left-avg( G ) ≤ m . (13)Indeed, since ⌊ log /q ( n ) ⌋ ≤ √ m ≤ m this is a direct consequence of Lemma 6with ν = and δ = 0.Finally, the lemma follows from (11), (12) and (13) if N is chosen to be atleast max( N , N , N ). q − √ m ≤ n ≤ q − m When the right side of the random bipartite graph becomes much larger thanthe left side, we cannot control the variance of S G anymore, and indeed Cheby-shev’s inequality cannot even give a positive probability, however small, thatthe graph contains many maximal stable sets of small left side. We thereforeuse another standard tool, Hoeffding’s inequality, which yields a tiny but non-zero probability. We then leverage this tiny probability to a high probabilityby cutting up the right side of the graph into a large number of large pieces toeach of which we apply Hoeffding’s theorem: Theorem 16 (Hoeffding [10]) . For i = 1 , . . . , s , let X i : Ω i → [0 , ρ ] be indepen-dent random variables, and let X = P si =1 X i . Then Pr[ X ≥ E[ X ] + λ ] ≤ e − λ sρ and Pr[ X ≤ E[ X ] − λ ] ≤ e − λ sρ for every λ with λ > . We will use Theorem 16 in the simpler case, when there is only one randomvariable, that is, s will be equal to 1.By S ′ G we denote the number of maximal stable sets S of G with | S ∩ L ( G ) | = ⌊ log /q ( ⌊ n/m log /q ( m ) ⌋ ) ⌋ . Note that, in contrast to S G , we put no restriction on | S ∩ R ( G ) | . Lemma 17.
For integers m, n let a ′ := ⌊ log /q ( ⌊ n/m log /q ( m ) ⌋ ) ⌋ and b := ⌊ log /q ( m ) ⌋ . Then there exists a constant c > so that for every ε > there isan N such that for G ∈ B ( m, n ; p )Pr (cid:20) S ′ G ≥ c (cid:18) ma ′ (cid:19) b − b (cid:21) ≥ − ε, for all m, n with m + n ≥ N and m /q ( m ) ≤ n ≤ q − m .Proof. In the following we assume that m /q ( m ) ≤ n ≤ q − m . Let k = m log /q ( m ) , and let R , R , . . . , R ⌊ k ⌋ be disjoint subsets of R ( G ) of size ⌊ n/k ⌋ G i = G [ L ( G ) ∪ R i ] for all 1 ≤ i ≤ ⌊ k ⌋ . Note that every G i may be considered as a random bipartite graph in B ( m, ⌊ n/k ⌋ ; p ).Let a ′ = ⌊ log /q ( ⌊ n/k ⌋ ) ⌋ and b = ⌊ log /q ( m ) ⌋ . Note that, since m /q ( m ) ≤ n ≤ q − m , also m ≥ a ′ and n ≥ b . Recall that, for 1 ≤ i ≤ k , S G i is the numberof maximal stable sets of G i with |S G i ∩ L ( G ) | = a ′ and |S G i ∩ R ( G ) | = b .Applying Lemma 13 to G i , we getE[ S G i ] ≥ c (cid:18) ma ′ (cid:19) b − b , (14)where c > S G i does never exceed (cid:0) ma ′ (cid:1) . Thus, by Theorem 16,Pr[ S G i ≤ E[ S G i ]] ≤ e − E [ S Gi ] ( ma ′ ) − (14) ≤ e − c b − b . The edge sets of the G i are pairwise disjoint, and thus the random variables S G i are independent:Pr[ S G i ≤ E [ S G i ] for all 1 ≤ i ≤ ⌊ k ⌋ ] ≤ e −⌊ k ⌋ c b − b . Since ⌊ k ⌋ = ⌊ m log /q ( m ) ⌋ , it dominates b − b . Hence, lim m →∞ ⌊ k ⌋ c b − b = ∞ . Thus, there is an N such that e −⌊ k ⌋ c b − b ≤ ε whenever m + n ≥ N , as m /q ( m ) ≤ n ≤ q − m implies that m grows with m + n . Hence, assuming m + n ≥ N ,Pr[ S G i ≤ E [ S G i ] for all 1 ≤ i ≤ ⌊ k ⌋ ] ≤ e −⌊ k ⌋ c b − b ≤ ε, (15)Thus, with probability 1 − ε , there is an i for which S G i ≥ c (cid:0) ma ′ (cid:1) b − b . Notethat every maximal stable set S of G i can be extended to a maximal stable set S ′ of G such that S ′ ∩ V ( G i ) = S . This extension is injective and, moreover, | S ′ ∩ L ( G ) | = | S ∩ L ( G ) | = a ′ . Hence, every maximal stable set counted by S G i is also counted by S ′ G , i.e., S ′ G ≥ S G i . This completes the proof.Observe that, in order to apply (15), we need k to dominate b b , where b = ⌊ log /q ( m ) ⌋ . Hence, k and thus also n should be of the order at least m /q ( m ) log /q (log /q ( m )) . This means that we could not use this method be-fore, when m and n had about the same size. Lemma 18.
For every ε > there is an N so that for G ∈ B ( m, n ; p )Pr (cid:2) left-avg( G ) ≤ m (cid:3) ≥ − ε, for all m, n with m + n ≥ N and q − √ m ≤ n ≤ q − m .Proof. In this proof consider integers m, n with q − √ m ≤ n ≤ q − m , and let a ′ := ⌊ log /q ( ⌊ n/m log /q ( m ) ⌋ ) ⌋ . Note that a ′ ≥ log /q ( n ) − log /q ( m log /q ( m ) ) − (cid:18) ma ′ (cid:19) ≥ (cid:16) ma ′ (cid:17) a ′ ≥ m log /q ( n ) ! a ′ ≥ m log /q ( n ) ! log /q ( n ) − log /q ( m log1 /q ( m ) ) − ≥ m log /q ( n ) − (log /q ( m ) +2) · (cid:16) log /q ( n ) (cid:17) − log /q ( n ) = n log /q ( m ) · m − (log /q ( m ) +2) · n − log /q (log /q ( n )) ≥ n log /q ( m ) · m − (log /q ( m ) +2) · n − log /q ( m/ = n log /q (16) · m − (log /q ( m ) +2) = n log /q (8) · n log /q (2) · m − (log /q ( m ) +2) In Lemma 17 the binomial coefficient (cid:0) ma ′ (cid:1) is divided by ⌊ log /q ( m ) ⌋ ⌊ log /q ( m ) ⌋ .So, let us compare this factor times m log /q ( m ) +2 against n log /q (2) . When m islarge enough, which we may assume since m /q ( m ) ≤ n ≤ q − m implies that m grows with m + n , we get that (log /q ( m )) ≥ /q (log /q ( m )). Thus m log /q ( m ) +2 · ⌊ log /q ( m ) ⌋ ⌊ log /q ( m ) ⌋ ≤ m log /q ( m ) +2 · (log /q ( m )) log /q ( m ) = m log /q ( m ) +2+log /q (log /q ( m )) ≤ m /q ( m ) . Using q − √ m ≤ n , we get m /q ( m ) ≤ (cid:16) (log /q ( n )) (cid:17) /q (log /q ( n ) )) = (cid:16) log /q ( n ) (cid:17) /q (log /q ( n )) = q − /q (log /q ( n )) . Since log /q (2) > n log /q (2) = q − log /q (2) · log /q ( n ) , we see that n log /q (2) >m /q ( m ) for large enough m and n .In conjunction with Lemma 17 this yields that there is a constant c > N so that Pr[ S ′ G ≥ cn log /q (8) ] ≥ − ε , (16)whenever m + n ≥ N .On the other hand, Lemma 11 yields an N so thatPr[ L G > n log q (1 / ] ≤ ε , when m + n ≥ N . 19ow we choose an N ≥ max( N , N ) so that 2 n log q (1 / = 2 n log /q (4) is muchsmaller than cn log /q (8) , by a factor of 2, say, when m + n ≥ N . (This is possibleas m /q ( m ) ≤ n ≤ q − m implies that m is large when m + n is large.) Thus2 S ′ G ≥ L G with a probability of ≥ − ε , and Lemma 6 (with δ = 0) completesthe proof. Indeed, note that S ′ G counts the number of maximal stable sets whoseleft sides have size ⌊ log /q ( ⌊ n/m log /q ( m ) ⌋ ) ⌋ ≤ log /q ( n ) ≤ m .Note that the estimation leading to (16) ceases to work when n > q − m . Weneed therefore a finer estimation, which is what we do in the next section. q − m ≤ n < q − m For κ ∈ (0 ,
1) the binary entropy is defined as H ( κ ) = κ log ( κ ) + (1 − κ ) log ( − κ ) . Observe that the binary entropy H ( κ ) is always strictly smaller than 1, exceptfor κ = . Moreover, H is monotonously increasing in the interval (cid:2) , (cid:3) . Forfurther details see [11].We will use the following bound on the binomial coefficient, which can befound for instance in Mitzenmacher and Upfal [11, Lemma 9.2]. Lemma 19.
For all m, k ∈ N with < k < m , (cid:18) mk (cid:19) ≥ m + 1 · H ( k/m ) · m . Lemma 20.
For integers m, n let λ := log /q ( n ) /m . For every ε, ϕ > thereis an N such that for G ∈ B ( m, n ; p ) , Pr[ S ′ G ≥ (1 − ϕ ) · H ( λ ) · m ] ≥ − ε, for all m, n with m + n ≥ N and q − m ≤ n ≤ q − m .Proof. Throughout the proof assume q − m ≤ n ≤ q − m .Let ϕ >
0. First we choose a δ with 0 < δ < H ((1 − δ ) κ ) ≥ (cid:16) − ϕ (cid:17) · H ( κ ) (17)for all κ ∈ (cid:2) , (cid:3) . This is possible since H is uniformly continuous in (cid:2) , (cid:3) and min { H ( κ ) : κ ∈ (cid:2) , (cid:3) } > a ′ := ⌊ log /q ( ⌊ n/m log /q ( m ) ⌋ ) ⌋ . Let N be such that when m + n ≥ N ⌈ (1 − δ ) log /q ( n ) ⌉ ≤ a ′ ≤ (cid:4) m (cid:5) . (18)The choice of N is possible since δ > /q ( n ) ≤ m . In the following,we restrict our attention to these m, n with m + n ≥ N . From (18) it followsthat (cid:18) ma ′ (cid:19) ≥ (cid:18) m ⌈ (1 − δ ) log /q ( n ) ⌉ (cid:19) . (19)20emma 19 gives (cid:18) m ⌈ (1 − δ ) log /q ( n ) ⌉ (cid:19) ≥ m + 1 · H ( ⌈ (1 − δ ) log /q ( n ) ⌉ m − ) · m . (20)Since H ( κ ) is monotonically increasing for κ ∈ (cid:2) , (cid:3) , it follows that H ( ⌈ (1 − δ ) log /q ( n ) ⌉ m − ) ≥ H ((1 − δ ) λ ), where we recall that λ = log /q ( n ) /m . As ≤ λ ≤ , we get2 H ( ⌈ (1 − δ ) log /q ( n ) ⌉ m − ) · m ≥ H ((1 − δ ) λ ) · m (17) ≥ − ϕ ) H ( λ ) · m . (21)Lemma 17 gives an N and a constant c > b = ⌊ log /q ( m ) ⌋ Pr (cid:20) S ′ G ≥ c (cid:18) ma ′ (cid:19) b − b (cid:21) ≥ − ε, when m + n ≥ N . Since q − m/ ≤ n ≤ q − m/ , there is an N such that m + n ≥ N implies cm + 1 · b − b · ϕ · H ( λ ) · m ≥ , where we use that λ ≥ . For such m and n , cm + 1 · b − b · (1 − ϕ ) · H ( λ ) · m ≥ (1 − ϕ ) · H ( λ ) · m . (22)Now, taking N = max( N , N , N ), the inequalities (19), (20), (21) and (22)finish the proof. Lemma 21.
For every α with ≤ α < and every ε > there is an N sothat for G ∈ B ( m, n ; p ) Pr (cid:2) left-avg( G ) ≤ m (cid:3) ≥ − ε, for all m, n with m + n ≥ N and q − m ≤ n ≤ q − αm .Proof. Let ≤ α < be given, and assume q − m ≤ n ≤ q − αm .Note that 2 κ < H ( κ ) for all ≤ κ ≤ α . Since H is continuous on thecompactum [ , α ], we may choose ϕ > ≤ κ ≤ α , 2 κ < (1 − ϕ ) H ( κ ). Moreover, we can put γ := min { (1 − ϕ ) H ( κ ) − κ : κ ∈ (cid:2) , α (cid:3) } and have γ > N such that m + n ≥ N yieldsPr[ S ′ G ≥ (1 − ϕ ) · H ( λ ) · m ] ≥ − ε , where λ = log /q ( n ) /m .By Lemma 11, there is an N such that Pr[ L G > /q ( n )+1 ] ≤ ε/ m + n ≥ N . Let N = max( N , N ) and assume m + n ≥ N . With probability1 − ε , S ′ G / L G ≥ (1 − ϕ ) · H ( λ ) · m · − /q ( n ) − ≥ γm − . Thus, there is an N ≥ N such that S ′ G / L G ≥ (1 − α ) − , whenever m + n ≥ N .Lemma 6 (with δ = 0 and ν = 1 − α ) completes the proof.21 .5 The case q − m ≤ n ≤ q − m Once the size n of the right side reaches q − m , the expected average left side ofa maximal stable set becomes very close to m , so close in fact that the methodsdeveloped so far begin to fail: We cannot any longer show that the average isat most m .The two main obstacles we face are: Firstly, when n approaches q − m/ theupper bound on L G , Lemma 11, becomes useless as it reaches 2 m . Secondly,the maximality requirement for maximal stable sets of small left side becomesharder to satisfy. Recall that we focus on left sides of size ≈ log /q ( n ) becausewith this size the maximality requirement eliminates only a constant proportionof the possible small maximal stable sets. However, when n surpasses q − m , themaximal stable sets with left side log /q ( n ) can no longer be considered as small ,since log /q ( n ) > m .Therefore we lower our goals and aim instead for an average of at most( + δ ) m , for any given δ >
0. Then the large maximal stable sets, those withleft side > ( + δ ) m , suddenly make up a significantly smaller proportion of thepower set.The key to that observation lies in the following basic lemma, a version ofwhich can be found in, for instance, van Lint [20, Theorem 1.4.5]. Lemma 22.
For all < γ < it holds that m X i = ⌈ γm ⌉ (cid:18) mi (cid:19) ≤ H (1 − γ ) m . Moreover, H (1 − γ ) < . Lemma 23.
For every δ > and ε > there is an N and an α < so thatfor G ∈ B ( m, n ; p ) Pr (cid:2) left-avg( G ) ≤ ( + δ ) m (cid:3) ≥ − ε, for all m, n with m + n ≥ N and q − αm ≤ n ≤ q − m .Proof. For G ∈ B ( m, n ; p ), let us denote by L δG the number of maximal stablesets S with | L ( G ) ∩ S | ≥ (cid:0) + δ (cid:1) m .Note that L δG ≤ m X i = ⌈ (1 / δ ) m ⌉ (cid:18) mi (cid:19) ≤ H (1 / − δ ) m , (23)where we used Lemma 22 for the second inequality.We show now that, with high probability, there are many more maximalstable sets with small left side in a random graph G ∈ B ( m, n ; p ), if q − αm ≤ n ≤ q − m and m + n ≥ N , for an N and an α < that we will determine below.In order to do so, note first that we may assume δ to be small enough so that α ′ := − δ ≥ . Next, fix n ′ ( m ) = n ′ := ⌈ q − α ′ m ⌉ and delete arbitrary n − n ′ R ( G ). The resulting graph G ′ may be viewed as a random graphin B ( m, n ′ ; p ), and we will see that with probability ≥ − ε it contains manymaximal stable sets with small left side. More precisely, we will prove that S ′ G ′ ≥ δ L δG (24)with probability at least 1 − ε . Note that the maximal stable sets counted by S ′ G ′ have a left side of size a ′ = ⌊ log /q ( ⌊ n ′ /m log /q ( m ) ⌋ ) ⌋ ≤ α ′ m = (cid:0) − δ (cid:1) m . Since every maximal stable set of G ′ extends to a maximal stable set of G withthe same left side, we may then use Lemma 6 with ν = δ to conclude thatleft-avg( G ) ≤ (cid:0) + δ (cid:1) m .Let us now see how we need to choose N and α in order to guarantee (24),which is all we need to finish the proof. For α , we could take α ′ if it were not forthe fact that we round up q − α ′ m to get n ′ (which turns out to be useful below).So we simply choose α to be somewhat larger than α ′ : Let α = − δ > α ′ and choose N large enough so that q − αm ≥ n ′ = ⌈ q − α ′ m ⌉ for all m, n with m + n ≥ N and n ≤ q − m . (This is possible as n ≤ q − m implies that m hasto be large as well if m + n is large.) Throughout the rest of the proof we willalways assume that m, n are integers with q − αm ≤ n ≤ q − m .Next, as H is monotonously increasing in the interval (cid:2) , (cid:3) , it follows that ϕ := 1 − H (cid:0) − δ (cid:1) H ( α ′ ) = 1 − H (cid:0) − δ (cid:1) H (cid:0) − δ (cid:1) < . Applying Lemma 20 with ε and ϕ yields an integer N ′ . Choose N ≥ N largeenough so that m + n ≥ N implies m + n ′ ≥ N ′ . (Again, this is possible as m has to be large if m + n is large.) Then, as α ′ ≥ , which in turn leads to n ′ ≥ q − m , we obtain for G ′ thatPr[ S ′ G ′ ≥ (1 − ϕ ) · H (log /q ( n ′ ) /m ) · m ] ≥ − ε, for all m with m + n ≥ N . Note that for m, n with m + n ≥ N H (log /q ( n ′ ) /m ) = H (log /q ( ⌈ q − α ′ m ⌉ ) /m ) ≥ H (log /q ( q − α ′ m ) /m ) = H ( α ′ )and thus 2 (1 − ϕ ) · H (log /q ( n ′ ) /m ) · m ≥ (1 − ϕ ) · H ( α ′ ) · m = 2 H (1 / − δ/ · m . Put µ := H (1 / − δ/ − H (1 / − δ ) and note that µ >
0. Inequality (23) givesthat with probability 1 − ε , S ′ G ′ / L δG ≥ ( H (1 / − δ/ − H (1 / − δ )) · m = 2 µm . Finally, choosing N ≥ N large enough so that 2 µm ≥ δ for all m, n with m + n ≥ N ensures (24). Again, this is possible as m grows with m + n .For the proof technique to work, we need G ′ to be a large graph. Otherwise,Lemma 20 cannot guarantee a high probability. In particular, m has to growwith m + n , which is why we assumed n ≤ q − m .23 .6 The case q − m ≤ n and proof of Theorem 4 If the right side of the random bipartite graph G is huge in comparision to theleft side, that is, if q − m ≤ n , then almost surely L ( G ) may be inductivelymatching into the right side. As a consequence, the set of left sides of maximalstable sets is equal to the power set of L ( G ), and thus left-avg( G ) = m . Lemma 24.
For every ε > there is an N so that for G ∈ B ( m, n ; p )Pr (cid:2) left-avg( G ) ≤ m (cid:3) ≥ − ε, for all m, n with m + n ≥ N and q − m ≤ n .Proof. Consider positive integers m, n with n ≥ q − m . We will give an N suchthat for all such m, n with m + n ≥ N there are, with probability 1 − ε , exactly2 m maximal stable sets in G . Then, every subset of L ( G ) is the left side of amaximal stable set, which implies left-avg( G ) = m .We proceed in a similar way as in the proof of Lemma 9 and therefore skipsome of the details. Let R , . . . , R ⌊ n/m ⌋ be pairwise disjoint subsets of R ( G ), ofsize m each. For i = 1 , . . . , ⌊ n/m ⌋ let M i be the random indicator variable foran induced matching of size m on L ( G ) ∪ R i . Since n ≥ q − m , n ≥ m and soPr [ M i = 1] ≥ p m q m . ThusPr ⌊ n/m ⌋ X i =1 M i = 0 ≤ (cid:16) − p m q m (cid:17) ⌊ n/m ⌋ ≤ e − p m q m ⌊ n/m ⌋ , where we use that 1 + x ≤ e x for all x ∈ R . Since n ≥ q − m , the term p m q m ⌊ n/m ⌋ becomes arbitrarily large for large m + n . Thus, there is an N sothat Pr hP ⌊ m/k ⌋ i =1 M i = 0 i ≤ ε for all m, n with m + n ≥ N and n ≥ q − m .Now assume that there is an induced matching on L ( G ) ∪ R i of size m forsome 1 ≤ i ≤ ⌊ n/m ⌋ . Then are 2 m many maximal stable sets of the graph G [ L ( G ) ∪ R i ] and each can be extended to a maximal stable set of G withoutchanging its left side. This completes the proof.Having exhausted all of the parameter space ( m, n ), we may finally proveour main theorem. Proof of Theorem 4.
For given ε > δ > N and α < asin Lemma 23. Then let N be at least as large as N and the N in Lem-mas 10, 15, 18, 21 (with α as chosen) and 24. Then the theorem follows. Acknowledgements
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Oliver Schaudt